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Yes. In fact, $\mathbb R^3$ could be any 3-manifold and $f(\mathbb R^1)$ any countable union of embedded segments.

Lemma 1. Let $U$ be an open ball in $\mathbb R^3$ and $f:I\to\mathbb R^3$ an embedding. Then $U\setminus f(I)$ is path-connected.

Proof. See Hatcher, Proposition 2B.1 on page 169. This proposition proves (among other things) that $S^3$ minus any embedded segment is path-connected. Extract the proof of this particular statement and repeat it word-by-word with $U$ in place of $S^3$. (The proof uses only the following facts about $U$: $U$ minus any closed sub-segment of $f(I)$ is open; $U$ minus any point of $\mathbb R^3$ is simply connected.)

Lemma 2. Let $f:I\to\mathbb R^3$ be an embedding and $s:I\to\mathbb R^3$ a path with endpoints outside $f(I)$. Then for any $\varepsilon>0$ there exists a path $s_\varepsilon:I\to\mathbb R^3$ which is $\varepsilon$-close to $s$ (in $C^0$), connects the same endpoints, and avoids $f(I)$.

Proof. Divide the domain of $s$ into intervals so small that the diameters of their images are less than $\varepsilon/10$. Let $p_0,p_1,\dots,p_n\in\mathbb R^3$ be the images of the division points ($p_0$ and $p_n$ are the endpoints of $s$). For each $p_i$, let $q_i$ be a point outside $f(I)$ such that $|p_i-q_i|<\varepsilon/10$ (for $i=0$ and $i=n$, choose $q_i=p_i$). Such a point $q_i$ exists since $f$ is injective and hence cannot cover a set with nonempty interior. Let $U_i$ be the ball of radius $\varepsilon/3$ centered at $p_i$. By Lemma 1, $U_i\setminus f(I)$ is path-connected. Therefore we can connect $q_i$ to $q_{i+1}$ by a path contained in $U_i$ and avoiding $f(I)$. The (suitably parametrized) product of these paths is the desired $s_\varepsilon$, q.e.d.

Now return to the original problem. The set $f(\mathbb R)$ is a countable union of sets $J_k$, $k\in\mathbb N$, where each $J_k$ is an injective image of a segment. We want to prove that any two points $p,q\in\mathbb R^3$ can be connected by a path avoiding $\bigcup J_k$.

Let $s_1$ be a path from $p$ to $q$ avoiding $J_1$ (such a path exists by Lemma 1). Let $\varepsilon_1$ be the minimum distance from $s_1$ to $J_1$, divided by 10. By Lemma 2, there exists a path $s_2$ which is $\varepsilon_1$-close to $s_1$ and avoids $J_1$. Let $\varepsilon_2$ be the minimum of $\varepsilon_1$ and the minimum distance from $s_1$ to $J_1$, divided by 10. Then there is a path $s_3$ which is $\varepsilon_2$-close to $s_2$ and avoids $J_2$, and so on. The resulting sequence $s_1,s_2,\dots$ converges (since $C^0$ is complete) to some path $s$ which avoids every set $J_k$.

Remark. The closure of $f(\mathbb R)$ can separate the space, for example, $f(\mathbb R)$ can be a dense subset of the torus.

Excellent answer: plenty of instructive technique on display here. Probably belongs on the Tricki somewhere. The invocation of completeness of a path space (and use of dependent choice) would seem to be unavoidable.
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Todd Trimble♦Nov 13 '10 at 12:03